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I was solving an exercise on power series and I ended up with the following limit for the radius:

$$ \lim_{n \to + \infty} \left( \frac{1}{n+1} + \ln \left(\frac{n+1}{n+2} \right) \right) n^2. $$

I tried to use the Taylor series for $\ln$ using just the first term and this gives me 1 as the result. But using some calculator I notice that the limit is actually $1/2$ and I found that this comes from the first two terms of the Taylor expansion. My question is: Why do I have to consider two terms? I think that this question can be consider simple and I hope to receive an answer anyway. Thank you

N230899
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  • How did you use the Taylor series for the logarithm? If $$\ln \biggl(\frac{n+1}{n+2}\biggr) = \ln \biggl(1 - \frac{1}{n+2}\biggr) = -\frac{1}{n+2} - \dotsb,,$$ the answer is that the second term of the Taylor series is of the same order as the sum of the first with the bit outside the logarithm. – Dermot Craddock Jun 30 '25 at 18:25
  • I'm sorry but I didn't get it yet – N230899 Jun 30 '25 at 19:49

3 Answers3

5

$$\ln\left(\frac{n+1}{n+2}\right)=-\ln\left(1+\frac{1}{n+1}\right)=\frac{-1}{n+1}+\frac{1}{2(n+1)^2}+o\left(\frac{1}{n^2}\right)$$ Therefore $$\frac{1}{n+1}+\ln\left(\frac{n+1}{n+2}\right)\sim\frac{1}{2n^2}$$

Christophe Boilley
  • 8,996
3

We know that $$ \ln\left(\frac{n+1}{n+2}\right) = \ln\left(1 - \frac{1}{n+2}\right). $$ Rembering that with Taylor's expansions, we have:

$$ \ln(1 - x) = -x - \frac{x^2}{2} + \text{smaller order terms}. $$

where $x = \frac{1}{n+2}$. Hence

$$ \frac{1}{n+1} + \ln\left(1 - \frac{1}{n+2}\right) \approx \frac{1}{n+1} - \frac{1}{n+2} - \frac{1}{2(n+2)^2}=$$$$\frac{(n+2) - (n+1)}{(n+1)(n+2)}- \frac{1}{2(n+2)^2} = \frac{1}{(n+1)(n+2)}- \frac{1}{2(n+2)^2}. $$

After,

$$ n^2 \left(\frac{1}{(n+1)(n+2)} - \frac{1}{2(n+2)^2}\right) = \frac{n^2}{(n+1)(n+2)} - \frac{n^2}{2(n+2)^2}. $$

\begin{align*} &\frac{n^2}{(n+1)(n+2)} = \frac{n^2}{n^2 + 3n + 2} \to 1, \\ &\frac{n^2}{2(n+2)^2} = \frac{1}{2} \cdot \frac{n^2}{n^2 + 4n + 4} \to \frac{1}{2}. \end{align*}

Definitively

$$ \lim_{n \to \infty} \left(\frac{1}{n+1} + \ln\left(\frac{n+1}{n+2}\right)\right) \cdot n^2 = 1 - \frac{1}{2} = \frac{1}{2}.$$

Sebastiano
  • 8,290
2

We have that

$$ \left( \frac{1}{n+1} + \ln \left(\frac{n+1}{n+2} \right) \right) n^2=$$

$$=\frac{n^2}{(n+2)^2}\frac{\frac{1}{n+2} + \ln \left(1-\frac{1}{n+2} \right)}{\frac1{(n+2)^2}}+\frac{n^2}{(n+1)(n-2)} \to 1\cdot\left(-\frac12\right)+1 =\frac12$$

using that by $x=\frac1{n+2} \to 0$

$$\frac{x+\log(1-x)}{x^2}\to -\frac12$$

which follows from $\log(1-x)=-x-\frac{x^2}2+o(x^2)$ or, as an alternative, using the result shown there

user
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